F2. Discrete random variables
Random variables
A random variable describes a random experiment by specifying the probability for every outcome in the outcome space.
Example: Role of an n-sided dice
\(X\) = “number of dots”
The probability for \(X=1\) is \(\frac{1}{n}\)
Curiosity - there is a dice that can be flipped like a coin The First Dice You Flip Like A Coin
Example: Blood sugar level
\(X\) = “blood sugar level (mmol/l) measured on an individual randomly chosen from a population”
Random variables - discrete and continuous
There are two types of random variables:
Discrete r.v. taking specific values such as numbers, natural numbers or categories
Continuous r.v. taking real numbers
Is the following r.v. discrete or continuous?
A toss of a coin resulting in heads or tail
The number of stars in the universe
The time (in seconds) it takes for the winner to reach the finish line in vasaloppet
The measured weight of an ant in milligram
From qualitative to quantitative description of an event
The outcome space for a toss of a coin is {heads, tails}
Let us define a discrete random variable for this random experiments by
\[X = \left\{ \begin{array}{lr} 1 & \text{if the outcome is heads}\\ 0 & \text{if the outcome is tails} \end{array}\right.\]
Tip
An outcome of the random variable \(X\) is denoted by a lowercase \(x\).
Discrete r.v. - probability function
A r.v. is defined with probabilities for the outcomes. This is for a discrete r.v. done with the probability function \(f_X(x) = P(X=x)\)
Tip
It is possible to simplify the description of the probability function by writing \(f(x)\).
The probability function can also be denoted \(p(x)\).
Discrete r.v. - Distribution function
All random variables can be described by a cumulative distribution function (CDF)
\(F_X(x) = P(X \leq x)\)
For a discrete r.v. the CDF is the sum of all probabilities for all outcomes that are less or equal than \(x\):
\(F_X(x) = \sum_{x_i\leq x} P(X=x_i) = \sum_{x_i\leq x} f_X(x_i)\)
The following holds
\(0\leq F_X(x) \leq 1\)
\(F_X(-\infty) = 0\)
\(F_X(\infty) = 1\)
Tip
It is possible to simplify the notation of the distribution function by writing \(F(x)\).
Probability distributions
A probability distribution (or just distribution) for a random variable makes it possible to calculated probabilities for outcomes or events.
Some probability distributions have names.
It is good to be familiar with common distributions.
Uniform discrete distribution
The classical dice
The random variable for tossing a six-sided dice
Probability function:
\(P(\text{the dice shows }x\text{ dots})=\frac{1}{6}\) where \(x=1,...,6\)
A different way to write it
\(f(x)=P(X=x)=\frac{1}{6}\) where \(x=1,...,6\)
Draw the distribution function!
What is the probability to get at least five dots?
What is the probability to get at most two dots?
What number of dots do we expect in average?
Simulation of 20 rolls with the dice
Simulation of 150 rolls with the dice
Poisson distribution
We use the Poisson distribution to describe how many times something is happening during a time interval
- The number of cars passing during an hour
The number of volcanic eruptions under 100 years
The number of birds you see when walking along a transect
Requirements for a Poisson distribution
In order for a random variable to follow a Poisson distribution the following must hold:
The average amount of events is the same for different time intervals
The number of events in non-overlapping time intervals are independent
Two events cannot occur at the same time
Volcanic eruptions
Note that the number of volcanic eruptions under 100 years might not comply with these requirements - which one?
Poisson distribution - Probability function
\(X \sim Po(\lambda)\)
(the symbol \(\sim\) (“tilde”) and what comes after is read out as “is distributed as a Poisson distribution”)
\(\lambda\) (greek letter “lambda”) is a parameter in the probability function
\(f_X(x) = P(X = x)=\frac{e^{-\lambda}\cdot \lambda^x}{x!}\)
where the outcome space consists of non-negative numbers \(x \in \{0,1,2,3,...\}\)
Poisson distribution
Example. Poisson distribution
A few years ago, a dermatologist raised an alarm that in an area in Lund, located near a chemical industry, the number of cases of a rare cancer disease was unusually high. In the particular area, nine people (six women and three men) had been affected by the disease over a five-year period. When the doctor studied the nationwide cancer registry, he saw that in a population similar to that in the particular area, one would have expected the number of cases to be four during this five-year period.
Model: \(X=\) “number of disease cases in the area under the five year period
\(X\sim Po(\lambda=4)\)
\(P(\text{exactly } x \text{ cases}) = P(X=x) = f(x) = \frac{e^{-4} 4^x}{x!}\)
What is the probability for exactly 5 cases?
What is the probability for at most 2 cases?
What is the probability for at least 9 cases?
Example. Poisson distribution
- What is the probability for exactly 5 cases?
Example. Poisson distribution
- What is the probability for at most 2 cases?
\(P(X\leq 2) = P(X=0) + P(X = 1) + P(X = 2)\)
Example. Poisson distribution
- What is the probability for at least 9 cases?
\(P(X\geq 9) = P(X=9) + P(X = 10) + ....\)
can also be written as
\(P(X\geq 9) = 1 - P(X \leq 8)\)
Derive the value on the probability using a probability table
A table for the distribution function for different parameter values
\(P(X \leq 8) = 0.97864\)
\(P(X \geq 9) = 1- P(X \leq 8) = 1 - 0.97864 = 0.02136\)
Binomial distribution
The binomial distribution exists when making \(n\) independent trails and then counting the number of times the trial was successful
Example. Binomial distribution
- You make 10 attempts to hit a paper bin with a paper ball
- You toss a coin 20 times and count how many times you got heads
Binomial distribution - an attempt to derive the formula for it’s probability function
We make \(n=10\) trials where the probability to succeed in one trial is \(p\)
\(X=\) “number of successful trials”
\(P(X=0) = P(\text{failur in all 10 trials}) = (1-p)(1-p)\cdots (1-p) = (1-p)^{10}\)
\(P(X=1)\)?
\(X=1\) correspond to one successful trial and nine failed trials, where the order does not matter
Success in the first trial: \(p(1-p)\dots (1-p) = p(1-p)^{9}\)
can happen in 10 ways, which gives \(P(X=1) = 10\cdot p(1-p)^{9}\)
By a similar reasoning: \(P(X=2) = \frac{10\cdot 9}{2}\cdot p^2(1-p)^8\)
Binomial distribution - probability function
\(X \sim Bin(n,p)\)
The parameters are \(n\), number of trials, and \(p\), the probability to succeed in a trial.
\(f_X(x) = P(X = x)== \frac{n!}{x!(n-x)!}p^x\cdot (1-p)^{n-x}\) where the outcome space consists of whole numbers between 0 and \(n\), i.e. \(x \in \{0,1,2,3,...,n\}\)
Binomial distribution
Example: We role a four-sided dice 7 times
\(X=\) “the number of times the dice shows one dot”
\(X \sim Bin(n,p)\) where \(n=7\) and \(p=\frac{1}{4}\)
- What is the probability to one dot 4 times?
\(P(X=4) = \binom{n}{x}p^x\cdot (1-p)^{n-x} = \binom{7}{4}\frac{1}{4}^4\cdot (1-\frac{1}{4})^{3}\)
- What is the probability to have one dot at the most four times?
\(P(X\leq 4) = \sum_{x = 0}^4 P(X=x) = \sum_{x = 0}^4 \binom{7}{x}\frac{1}{4}^{x}\cdot (\frac{3}{4})^{7-x}\)
Tedious to calculate
Derive the value on the probability using a probability table
A table for the distribution function for different parameter values
\(P(X\leq 4) = 0.98712\)
Expected value and variance
There are several ways to summarise the random variation for a random variable \(X\)
Measure of central tendency:
Expected value ”the value that \(X\) takes in average”
\(E(X)\)
\(\mu\)
Measure of spread:
Variance - ”how the values of \(X\) spread around the expected value”
\(V(X)\)
\(\sigma^2\)
Standard deviation - \(\sqrt{V(X)}\)
\(\sigma\)
Tip
All random variables have expected values and variance denoted as \(\mu\) and \(\sigma^2\) no matter what probability distribution they belong to.
E.g. the expected value and variance for r.v. \(X\) is \(\mu_X\) and \(\sigma^2_X\)
The expected value and variance for r.v. \(Y\) is \(\mu_Y\) and \(\sigma^2_Y\)
Expected value for a discrete random variable
\(E(X) = \sum_{\text{all }x} x\cdot P(X=x) = \sum_{\text{all }x} x\cdot f(x)\)
Expected value for a discrete uniform distribution
\(E(X)=\sum_{x=1}^6 x\cdot \frac{1}{6} = \frac{1}{6} \cdot (1 + 2+ 3 +4+5+6) = 3.5\)
Fits well to the simulations we made earlier
Expected value for a Poisson distribution
\(X \sim Po(\lambda)\)
\(E(X) = \lambda\)
In the formula sheet the parameter for the Poisson distribution is \(\mu\)
Expected value for a Binomial distribution
\(X \sim Bin(n,p)\)
\(E(X) = n\cdot p\)
Expected value of a function of a random variable
\(g(x)\) is a function
\(E(g(X)) = \sum_{\text{all }x} g(x)\cdot f(x)\)
the probability function does not change when values are transformed